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  • #46
    trust me

    it's the square root of ratio of masses

    KE = 0.5 * 12 * ( v1 )^2

    KE = 0.5 * 16 * ( v2 )^2

    divide equations by 1 another ->

    1 = (12/16) * (v1)^2/(v2)^2 ->

    (v2/v1)^2 = 12/16 ->

    v2/v1 = ( 12/16)^0.5

    distance increase is proportional to velocity increase ( 1st order relationship )

    it works the same for runners who lose blubber but not muscle ( for a small range of values )

    i e a 10.00 guy who weighs 77kg, but ideal muscular weight is 75kg, can be capable of

    10.00 * ( 75/77 )^1/2 = 9.87

    or if you have a 3'32 guy who weighs 67kg,but shoud be ideally 65kg

    -> 3'32 * ( 65/67 )^1/2 = 3'28.9

    the question is when weight loss bites into your muscle & power loss...

    Comment


    • #47
      Originally posted by eldrick
      trust me

      it's the square root of ratio of masses

      KE = 0.5 * 12 * ( v1 )^2

      KE = 0.5 * 16 * ( v2 )^2

      divide equations by 1 another ->

      1 = (12/16) * (v1)^2/(v2)^2 ->

      (v2/v1)^2 = 12/16 ->

      v2/v1 = ( 12/16)^0.5
      Everything so far makes sense.

      distance increase is proportional to velocity increase ( 1st order relationship )
      That isn't true. The distance a projectile travels (until it reaches the same height again) is given by

      d=2(v^2)(sinx)(cosx) / g, where x is the release angle, and g is gravity.

      Proof:
      The vertical component of the velocity is vsinx
      The horizontal component of the velocity is vcosx

      The amount of time spent in the air is twice the time it takes the object to stop traveling upwards
      t=2*vsinx/g
      The distance traveled horizontally is velocity*time, or
      d=(vcosx)(2vsinx/g), or
      d=2(v^2)(sinx)(cosx)/g

      Comment


      • #48
        Originally posted by rainy.here
        Originally posted by eldrick
        trust me
        it's the square root of ratio of masses
        KE = 0.5 * 12 * ( v1 )^2
        KE = 0.5 * 16 * ( v2 )^2
        1 = (12/16) * (v1)^2/(v2)^2 ->
        (v2/v1)^2 = 12/16 ->
        v2/v1 = ( 12/16)^0.5
        Everything so far makes sense.
        distance increase is proportional to velocity increase ( 1st order relationship )
        That isn't true. The distance a projectile travels (until it reaches the same height again) is given by
        d=2(v^2)(sinx)(cosx) / g, where x is the release angle, and g is gravity.
        Proof:
        The vertical component of the velocity is vsinx
        The horizontal component of the velocity is vcosx
        The amount of time spent in the air is twice the time it takes the object to stop traveling upwards
        t=2*vsinx/g
        The distance traveled horizontally is velocity*time, or
        d=(vcosx)(2vsinx/g), or
        d=2(v^2)(sinx)(cosx)/g
        OMG :shock: It's spawning!!

        Comment


        • #49
          Originally posted by Marlow

          OMG :shock: It's spawning!!
          Battle of the brainiacs, there's gonna be grey matter everywhere!
          phsstt!

          Comment


          • #50
            Originally posted by rainy.here
            That isn't true. The distance a projectile travels (until it reaches the same height again) is given by

            d=2(v^2)(sinx)(cosx) / g, where x is the release angle, and g is gravity.

            Proof:
            The vertical component of the velocity is vsinx
            The horizontal component of the velocity is vcosx

            The amount of time spent in the air is twice the time it takes the object to stop traveling upwards
            t=2*vsinx/g
            The distance traveled horizontally is velocity*time, or
            d=(vcosx)(2vsinx/g), or
            d=2(v^2)(sinx)(cosx)/g
            thanks for correction

            i shouda looked up some ole posts

            http://mb.trackandfieldnews.com/discuss ... 993#127993

            but i didn't quote the basic range equation to validate it - not worth bothering with

            anyhows, small correction, the basic range equation quoted is for same level height of release/landing projectiles - the case for projectiles released from a height is fiendishly difficult to work out accurately ( why it's better to use projectile software, but you'll get very similar results with basic range formula )

            http://mb.trackandfieldnews.com/discuss ... 099#307099

            that was my "go to" site, but unfortunately he's dumbed it down & removed all the good stuff ( i imagine it's gone to subscription-only access ) - he stops at (4) when the meaty stuff went upto equation (12), which gave proper formula

            Comment


            • #51
              Originally posted by eldrick
              but i didn't quote the basic range equation to validate it - not worth bothering with

              anyhows, small correction, the basic range equation quoted is for same level height of release/landing projectiles - the case for projectiles released from a height is fiendishly difficult to work out accurately ( why it's better to use projectile software, but you'll get very similar results with basic range formula )

              http://mb.trackandfieldnews.com/discuss ... 099#307099

              that was my "go to" site, but unfortunately he's dumbed it down & removed all the good stuff ( i imagine it's gone to subscription-only access ) - he stops at (4) when the meaty stuff went upto equation (12), which gave proper formula
              Which "good stuff" are you referring to? There is a good chance I can derive the equations, unless it involves air resistance. Being the geek I am, this stuff is interesting to me and a whole lot more interesting than my job.

              Comment


              • #52
                Originally posted by rainy.here
                Originally posted by eldrick
                trust me

                it's the square root of ratio of masses

                KE = 0.5 * 12 * ( v1 )^2

                KE = 0.5 * 16 * ( v2 )^2

                divide equations by 1 another ->

                1 = (12/16) * (v1)^2/(v2)^2 ->

                (v2/v1)^2 = 12/16 ->

                v2/v1 = ( 12/16)^0.5
                Everything so far makes sense.

                distance increase is proportional to velocity increase ( 1st order relationship )
                That isn't true. The distance a projectile travels (until it reaches the same height again) is given by

                d=2(v^2)(sinx)(cosx) / g, where x is the release angle, and g is gravity.

                Proof:
                The vertical component of the velocity is vsinx
                The horizontal component of the velocity is vcosx

                The amount of time spent in the air is twice the time it takes the object to stop traveling upwards
                t=2*vsinx/g
                The distance traveled horizontally is velocity*time, or
                d=(vcosx)(2vsinx/g), or
                d=2(v^2)(sinx)(cosx)/g
                Those equations rainy is quoting are only for without air resistance. In college I did an independent study on the mathematical equations of golf ball flight, which is similar, but more complicated because of the lift caused by spin. The equations with air resistance are related to the hyperbolic inverse tangent. I would guess this is fairly negligible for a shot however. Certainly one thrown by me!

                Comment


                • #53
                  Originally posted by rainy.here
                  Which "good stuff" are you referring to? There is a good chance I can derive the equations, unless it involves air resistance. Being the geek I am, this stuff is interesting to me and a whole lot more interesting than my job.
                  the equations (5) to (12) listed as existing in a 2 1/2y ole post which do not exist on that site anymore

                  try it :

                  range equation with projectile released from height h at angle A

                  ( if you can derive max range/angle off it, that is special )

                  Comment


                  • #54
                    actually, save above to later

                    i always use this to gauge the quality of a numbercruncher :

                    i'll post you the best puzzle i ever saw ( - it involves, knowledge of maths, logic & commonsense - if you can solve it, i believe your mind is sharp enough, resourceful enough & functioning well enough to solve most any situation you'll ever encounter )

                    it's from another book & here's the puzzle :

                    "The Golden Ball & the Oxford Professors :

                    An Oxford Professor named Hall
                    Possessesed an Octagonal Ball,
                    The Square of it's Weight
                    Divided by Eight
                    Was Pi times the Root of Sod All.

                    Now the more perceptive reader will have spotted immediately the logical & mathematical flaws in the above statement. If it was octagonal, it can hardly have been a ball & the maths seems to contain serious flaws. I am now in a position to reveal the true story. The ball was in fact a ball, not octagonal , & made of pure gold. Together with another Professor he had obtained it in illicit fashion during an archaeological dig in peru. The 2 accomplices being born complexifiers, fell into a dispute about how they should divide up this valuable object.

                    One had a fancy to have a solid gold paperweight & as a ball is not much use for that purpose, decided that it must be a cylinder ; so he said,

                    " All I want is a cylinder from the ball & I can turn this up on the lathe in the laboratory. All the rest of it, the golden swarf, you shall have & you can sell it for a considerable sum. "

                    The 2nd Professor did some calculations & proved to his own satisfaction that any true cylinder from the sphere must contain less than half the volume of the sphere & so he agreed to the terms.

                    Was he wise ?

                    If the total weight of the of gold in the ball was 1 kg, what was the least weight of swarf his friend could make in turning a true cylinder from the golden ball ? "


                    N.B. You need to look up the formulas of the volumes of a sphere & a cylinder - it's in wiki

                    A virtual 5 pints to anyone who can solve it ( JRM is initially banned from entering a solution ( because it's meant for us laymen to have some fun ), but after a coupla days, it would be nice for him to post the most "elegant" solution for us )
                    solve that & you are doing as well as any 16y ole good maths student

                    next/last puzzle after, is the horrendous "ladder puzzle" ( post after you solve above )

                    Comment


                    • #55
                      Originally posted by bambam
                      The equations with air resistance are related to the hyperbolic inverse tangent
                      very special abilty

                      i see now how you got to 38th

                      you never struck me as a numbercruncher, but doing that outgeeks me, as at school, that was in "advanced" maths syallabus & for me to do that i'd have to drop chemistry, which for some unknown reason was most widely needed science subject - only subject mandatory for any med school applicants ( not biology ( ?! ), which i dropped at 16y )

                      Comment


                      • #56
                        Originally posted by eldrick
                        Originally posted by bambam
                        The equations with air resistance are related to the hyperbolic inverse tangent
                        very special abilty

                        i see now how you got to 38th

                        you never struck me as a numbercruncher, but doing that outgeeks me, as at school, that was in "advanced" maths syallabus & for me to do that i'd have to drop chemistry, which for some unknown reason was most widely needed science subject - only subject mandatory for any med school applicants ( not biology ( ?! ), which i dropped at 16y )
                        Math major in college - loved math - still read popular books about it - Fermat's Theorem, Riemann Hypothesis. Good stuff. I buy them at B&N with my wife, and she usually says something like "You must be the only person who has ever bought that book!"

                        Comment


                        • #57
                          i've got the 2 fermat books - singh then azcel, but one on reimann ??

                          please post any info as i'll have to order it immediately

                          Comment


                          • #58
                            Originally posted by eldrick

                            If the total weight of the of gold in the ball was 1 kg, what was the least weight of swarf his friend could make in turning a true cylinder from the golden ball ?

                            N.B. You need to look up the formulas of the volumes of a sphere & a cylinder - it's in wiki

                            A virtual 5 pints to anyone who can solve it ( JRM is initially banned from entering a solution ( because it's meant for us laymen to have some fun ), but after a coupla days, it would be nice for him to post the most "elegant" solution for us )

                            solve that & you are doing as well as any 16y ole good maths student

                            next/last puzzle after, is the horrendous "ladder puzzle" ( post after you solve above )
                            I suspect that anyone who can solve that already knows (or can derive) the formulae for the volume of spheres and cylinders. My solution to the above is probably the "standard" calculus response. I'll have to think of a better way of doing it when I get some time.

                            Assume our sphere has radius 1. Let r represent the radius of the cylinder we're creating.

                            Then:
                            V_sphere = 4Pi/3
                            V_cylinder = Pi r^2 h
                            (1) V_cylinder = 2 Pi r^2 Sqrt(1-r^2) (pythagoras to find half the height of the cylinder)

                            We want the largest value of V_c / V_s
                            V_c / V_s = [ 2 Pi r^2 Sqrt(1-r^2) ] / [4Pi/3]
                            (2) = 3/2 r^2 Sqrt(1-r^2)

                            The 3/2 is irrelevant, so we want to maximize the value of r^2 Sqrt(1-r^2).
                            It's considerably easier to maximize the value of x = r^4 (1-r^2) instead, and doesn't change anything.

                            x' = 4r^3 - 6r^5
                            gives r = 0, +- Sqrt(2/3), and obviously r = Sqrt(2/3) is the value of r we want.

                            Plugging this value of r back into equation (1) above, gives a cylinder volume of
                            V_cylinder = 4Pi/3Sqrt(3), and plugging this into equation (2) above tells us the relative volume/mass of the cylinder to the sphere.

                            V_cylinder / V_sphere = (3/2) * (2/3) * Sqrt(1-2/3)
                            which = 1/sqrt(3)

                            Comment


                            • #59
                              very good

                              try the last one ( you can try deriving range from for a projectile from an elevated height & optimum angle later )

                              "2 ladders 20 & 30 feet long, lean in opposite directions across a passageway. They cross at a point 8 feet above the floor. How wide is the passage ?"

                              Comment


                              • #60
                                http://www.physicsforums.com/archive/in ... 19674.html

                                You've been around the block before.

                                Comment

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